Difference between revisions of "2013 AMC 12A Problems/Problem 14"

Revision as of 18:41, 22 February 2013

The sequence

$\log_{12}{162}$ , $\log_{12}{x}$ , $\log_{12}{y}$ , $\log_{12}{z}$ , $\log_{12}{1250}$

is an arithmetic progression. What is $x$ ?

$\textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}$

Since the sequence is arithmetic,

$\log_{12}{162}$ + $4d$ = $\log_{12}{1250}$ , where $d$ is the common difference.

Therefore,

$4d$ = $\log_{12}{1250}$ - $\log_{12}{162}$ = $\log_{12}{(1250/162)}$ , and

$d$ = $\frac{1}{4}$ ( $\log_{12}{(1250/162)}$ ) = $\log_{12}{(1250/162)^{1/4}}$

Now that we found $d$ , we just add it to the first term to find $x$ :

$\log_{12}{162}$ + $\log_{12}{(1250/162)^{1/4}}$ = $\log_{12}{((162)(1250/162)^{1/4})}$

$x$ = $(162)$ $(1250/162)^{1/4}$ = $(162)$ $(625/81)^{1/4}$ = $(162)(5/3)$ = $270$ , which is $B$

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 13	Followed by Problem 15
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions

@@ Line 1: / Line 1: @@
+== Problem==
+The sequence
+<math>\log_{12}{162}</math>, <math>\log_{12}{x}</math>, <math>\log_{12}{y}</math>, <math>\log_{12}{z}</math>, <math>\log_{12}{1250}</math>
+is an arithmetic progression. What is <math>x</math>?
+<math> \textbf{(A)} \ 125\sqrt{3} \qquad \textbf{(B)} \ 270 \qquad \textbf{(C)} \ 162\sqrt{5} \qquad \textbf{(D)} \ 434 \qquad \textbf{(E)} \ 225\sqrt{6}</math>
+==Solution==
 Since the sequence is arithmetic,
@@ Line 17: / Line 29: @@
 <math>x</math> = <math>(162)</math><math>(1250/162)^{1/4}</math> = <math>(162)</math><math>(625/81)^{1/4}</math> = <math>(162)(5/3)</math> = <math>270</math>, which is <math>B</math>
+== See also ==
+{{AMC12 box|year=2013|ab=A|num-b=13|num-a=15}}